# Kepler’s laws simplified: explanation, exercises, experiment

Kepler ‘s **laws** of planetary motion were made by the German astronomer Johannes Kepler (1571-1630). Kepler deduced them based on the work of his teacher, the Danish astronomer Tycho Brahe (1546-1601).

Brahe has carefully compiled data on planetary motions for over 20 years, with surprising precision and accuracy, considering that the telescope had not yet been invented at the time. The validity of your data remains in effect today.

**Kepler’s 3 Laws**

Kepler’s laws state:

**First law** : all planets describe elliptical orbits with the Sun at one of the focuses.

*– ***Second law or law of equal areas:** a direct line from the Sun to any planet (focal ray) sweeps equal areas at equal times.

*– ***Third law:** The square of time it takes for any planet to orbit the Sun is proportional to the cube of its average distance from the Sun.

Let *T be the* referred time, called *the orbital period* , and *r be* the average distance, then:

*T ^{2} is proportional to ^{3}*

*T = kr ^{3}*

This means that the ratio of *T ^{2} / r ^{3}* is the same for all planets, which makes it possible to calculate the orbital radius, if the orbital period is known.

When *T* is expressed in years and *r* in astronomical units UA *, the proportionality constant is k = 1:

*T ^{2} = R ^{3}*

* One astronomical unit is equal to 150 million km, which is the average distance between the Earth and the Sun. The Earth’s orbital period is 1 year.

__Universal law of gravitation and Kepler’s third law__

__Universal law of gravitation and Kepler’s third law__

The law of universal gravitation states that the magnitude of the gravitational force of attraction between two objects of mass *M* and *m* , respectively, whose centers are separated by a distance *r,* is given by:

*F = G mM / r ^{2}*

G is the universal gravitational constant and its value is G = 6,674 x 10 ^{-11} Nm ^{2} / kg ^{2} .

Now, the planets’ orbits are elliptical with a very small eccentricity.

This means that the orbit is not far from a circumference, except in some cases like the dwarf planet Pluto. If we approximate the orbits to the circular shape, the acceleration of the planet’s motion is:

*a _{c} = v ^{2} / r*

Since *F = ma* , we have:

*G mM / r ^{2} = mv ^{2} / r*

Here *v* is the linear velocity of the planet around the Sun, static and mass assumption *M* , while that of the planet is *m* . Thus:

This explains that planets farther away from the Sun have a lower orbital speed, as this depends on *1 / √r* .

Since the distance the planet travels is approximately the length of the circumference: L = 2πr and it takes a time equal to T, the orbital period, to obtain:

*v = 2πr / T*

Equating the two expressions for v, we obtain a valid expression for T ^{2} , the square of the orbital period:

And it is precisely Kepler’s third law, since in this expression the *4π ^{2} / GM* parenthesis is constant, so

*T*is proportional to the distance

^{2}*r*cubed.

The final equation for the orbital period is obtained by extracting the square root:

How much is the mass of the Sun worth? It is possible to find out using this equation. We know that the Earth’s orbital period is one year and the orbital radius is 1 AU, equivalent to 150 million kilometers, so we have all the necessary data.

In our previous equation, we solve for *M* , but not before converting all values to the International System of SI Units:

1 year = 3.16 x 10 ^{7} seconds.

1 AU = 150 million km = 1.5 x10 ^{11} m.

**Exercises**

Although Kepler only had the planets in mind when he derived his famous laws, they are also valid for the movement of satellites and other bodies in the solar system, as we will see below.

**– Exercise 1**

Knowing that Jupiter’s orbit is 5.19 times larger than Earth’s, find Jupiter’s orbital period.

**Solution**

According to the definition of the Astronomical Unit, Jupiter is distant from the Sun 5.19 AU, therefore, according to Kepler’s third law:

*T ^{2} = R ^{3} = (5.19) ^{3} years*

Therefore *T = (5.19) ^{3/2} years = 11.8 years*

**– Exercise 2**

Comet Halley visits the Sun every 75.3 years. Meet:

a) The semi-main axis of its orbit.

b) The aphelion measurement, if the perihelion measures 0.568 AU.

**Solution**

Comet Halley visits the Sun every 75.3 years. Meet:

a) The semi-main axis of its orbit.

b) The aphelion measurement, if the perihelion measures 0.568 AU.

**Solution for**

When a planet or any other star is at the closest point to the Sun, it is said to be at *perihelion,* and when it is farther away, at *aphelion* . In the special case of a circular orbit, r in Kepler’s third law is the radius of the orbit.

However, in the elliptical orbit, the celestial body is more or less distant from the Sun, with the semi-major axis “a” being the mean between aphelion and perihelion:

Therefore, we substitute r for a in Kepler’s third law, which results in Halley in:

*T ^{2} = a ^{3} → a = (T) ^{2/3} → a = (75.3) ^{2/3} AU = 17,832 AU*

**Solution b**

a = ½ (Perihelion + Aphelion)

*17,832 = ½ (0.568+ Aphelion) → Aphelion = 2 x 17,832 – 0.568 AU = 35.10 AU.*

__To experiment__

__To experiment__

Analyzing the movement of planets requires weeks, months and even years of careful observation and recording. But a very simple-scale experiment can be performed in the laboratory to prove that Kepler’s law of equal areas holds.

For this, a physical system in which the force that governs the movement is central is necessary, a sufficient condition for the law of areas to be fulfilled. This system consists of a mass tied to a long cable, with the other end of the wire attached to a support.

The mass is moved a small angle away from its equilibrium position and a slight thrust is imprinted on it so that it performs an oval (almost elliptical) motion in the horizontal plane, as if it were a planet around the Sun.

In the curve described by the pendulum, we can prove that it sweeps equal areas in equal times, if:

– We consider vector rays that go from the center of attraction (initial equilibrium point) to the position of the mass.

-And we sweep between two consecutive moments of equal duration, in two different areas of movement.

The longer the pendulum thread and the smaller the angle of the vertical, the net restorative force will be more horizontal and the simulation is similar to the case of moving with central force in a plane.

Then, the described oval approaches an ellipse, like the one traveled by planets.

**materials**** **

– Extendable thread

-1 white painted metal mass or ball that works as a pendulum lentil

-Ruler

-Conveyor

-Camera with automatic strobe disc

-Supports

-Two light sources

-A sheet of black paper or cardboard

**Process**

Assembling the figure is necessary to take several intermittent shots of the pendulum as it goes its way. To do this, you must place the camera just above the pendulum and the automatic strobe disc in front of the lens.

In this way, images are obtained at regular intervals from the pendulum, for example, every 0.1 or every 0.2 seconds, which allows us to know the time needed to move from one point to another.

The pendulum mass must also be properly lit by placing the lights on both sides. Duckweed should be painted white to improve contrast on the background, which consists of black paper spread on the floor.

Now you must check that the pendulum sweeps equal areas at equal times. For this, a time interval is chosen and the points occupied by the pendulum in that interval are marked on the paper.

In the image, a line is drawn from the center of the oval to these points and therefore we have the first of the areas swept by the pendulum, which is approximately an elliptical sector like the one shown below:

**Calculating the area of the elliptical section**

Angles *θ _{o}* and

*θ*are measured with the protractor , and this formula is used to find S, the area of the elliptical sector:

_{1}*S = F (θ _{1} ) – F (θ _{o} )*

With *F (θ)* given by:

Note that *a* and *b* are the largest and smallest semi-axis, respectively. The reader only has to worry about carefully measuring the axes and angles, as there are online calculators to easily evaluate this expression.

However, if you insist on doing the calculation manually, remember that the angle θ is measured in degrees, but when entering data into the calculator, the values must be in radians.

Then it is necessary to mark another pair of points in which the pendulum has invested the same time interval and draw the corresponding area, calculating its value with the same procedure.

**Checking the law of equal areas**

Finally, it remains to be verified whether the law of areas is fulfilled, that is, that equal areas are swept at equal times.

Are the results deviating a little from what was expected? It should always be remembered that all measurements are accompanied by their respective experimental errors.